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The next set of medium/hard DS number properties questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers.

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number (2) \(2\sqrt{x^2}\) is a prime number.

2. If a positive integer n has exactly two positive factors what is the value of n?

Notice that, n has exactly two positive factors simply means that n is a prime number, so its factors are 1 and n itself.

(1) n/2 is one of the factors of n. Since n/2 cannot equal to n, then n/2=1, thus n=2. Sufficient.

(2) The lowest common multiple of n and n + 10 is an even number. If n is an odd prime, then n+10 is also odd. The LCM of two odd numbers cannot be even, therefore n is an even prime, so 2. Sufficient.

Answer: D.

Dear Bunuel

That is a great observation: The LCM of two odd numbers cannot be even

Can we take it as a general observation and expand it further, Like:

1) The LCM of two ODD numbers will always be ODD. 2) The LCM of two EVEN numbers will always be EVEN.

Or further expand it Like: 3) The LCM of any number of ODD Integers will always be ODD. 4) The LCM of any number of EVEN Integers will always be EVEN.

I guess this bit of Info can be great aid on the Test day.

Hi, In question 3 above, it says that X and Y have 3 postive factors. But it does not say only positive factors. Could the number not have other factors such as -3,-1,-9 apart from 3,1,9? What am I missing out? It will be great if someone can point what I am doing incorrectly. Thanks!

Hi, In question 3 above, it says that X and Y have 3 postive factors. But it does not say only positive factors. Could the number not have other factors such as -3,-1,-9 apart from 3,1,9? What am I missing out? It will be great if someone can point what I am doing incorrectly. Thanks!

In GMAT, factors are always positive.

On the other hand, multiples can be positive as well as negative.
_________________

2. If a positive integer n has exactly two positive factors what is the value of n?

(1) n/2 is one of the factors of n (2) The lowest common multiple of n and n + 10 is an even number.

As per Question Stem, 'n' has exactly 2 positive factors, it has to be prime. i.e. 1 & itself

1) As per this, 'n' has to be 2 SUFF

2) As per this, 'n' has to be even. Referring to Question stem, it has to be 2 SUFF

Ans : D

Please correct me if I'm wrong.
_________________

Cheers, BG

Hit 'Kudos' if you find my post helpful

OG 13 Verbal Directory: http://gmatclub.com/forum/official-guide-verbal-question-directory-og-13-and-og-161949.html#p1121171 Question Of the Day: http://gmatclub.com/blog/category/blog/gmat-tests/?fl=menu Extreme Collection of 2000 PS & DS: http://gmatclub.com/forum/1001-ds-questions-file-106193.html?fl=similar CR Tips & Approach to arrive at Best Answer: http://gmatclub.com/forum/cr-methods-an-approach-to-find-the-best-answers-93146.html

Arise, Awake and Stop not till the goal is reached. I won't stop till I hit 700+

The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).

Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.

(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.

Answer: A.

first and foremost, thanks for this problem set. it's helped me immensely

Can you offer more explanation on St. (1) and St. (2), specifically:

St. (1): which portion of the equation represents any specific "power" that is distributed to 2 ? All I can gather is that 2 will be multiplied by 3 (odd) and then multiplied by some odd number (2q + 1). In effect you are multiplying an even (6) by some odd number. How does this tell us the power that 2 is raised to?

Example:

x = 2*3*[2*2 +1) = 18

Prime factors of 18 = 2^1 and 3^2. How does the equation (2q+1) ensure that 2 will always be raised to an odd power?

St. (2): Does it make sense/is it possible to break it out in similar fashion as you did to statement 1? x = 2(7p+1) ? The (7p+1) could yield an odd or even number, depending on what q is (as displayed in your example).

Also, what program do you use for equation writing?

The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).

Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.

(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.

Answer: A.

first and foremost, thanks for this problem set. it's helped me immensely

Can you offer more explanation on St. (1) and St. (2), specifically:

St. (1): which portion of the equation represents any specific "power" that is distributed to 2 ? All I can gather is that 2 will be multiplied by 3 (odd) and then multiplied by some odd number (2q + 1). In effect you are multiplying an even (6) by some odd number. How does this tell us the power that 2 is raised to?

Example:

x = 2*3*[2*2 +1) = 18

Prime factors of 18 = 2^1 and 3^2. How does the equation (2q+1) ensure that 2 will always be raised to an odd power?

St. (2): Does it make sense/is it possible to break it out in similar fashion as you did to statement 1? x = 2(7p+1) ? The (7p+1) could yield an odd or even number, depending on what q is (as displayed in your example).

Also, what program do you use for equation writing?

For (1): \(x=2*3*(2q+1)=2*odd*odd\) --> two odd multiples there obviously do not have 2 as their factors, thus x has 2 in first power only (we have only one 2 there).

Is Statement 2 something we can expect to see on the GMAT ?

Yes, you do have only positive roots on GMAT. \(\sqrt{4} = 2\) only (not -2) \(\sqrt{x}\) is never negative

Bot note here that the expression is different: \(\sqrt{x^2}\) \(\sqrt{x^2} = |x|\) (absolute value of x) x can be 1 or -1 here because it is squared. \(\sqrt{1}\) will be 1 only.

So, if \(x^2 = 4\), x = 2 or -2 But \(\sqrt{4} = 2\) only on the GMAT
_________________

8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Answer: B.

Theory: Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.

Hi Bunuel Quick question on Statement 1 Here Let us assume SET S with 10 elements and let 5 th and 6th element be 1/a and 1/b where a and b are both prime. Now the median of the set would be a+b/2ab and it is given that reciprocal of the median is also prime.which means that 2(ab/a+b) will also be prime. So we have to achieve this ab=a+b and that is only possible when a and b are equal to 2.And this proves that Statement 1 is sufficient to answer the question. Can you please point where did I went wrong.

8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Answer: B.

Theory: Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.

Hi Bunuel Quick question on Statement 1 Here Let us assume SET S with 10 elements and let 5 th and 6th element be 1/a and 1/b where a and b are both prime. Now the median of the set would be a+b/2ab and it is given that reciprocal of the median is also prime.which means that 2(ab/a+b) will also be prime. So we have to achieve this ab=a+b and that is only possible when a and b are equal to 2.And this proves that Statement 1 is sufficient to answer the question. Can you please point where did I went wrong.

2ab/(a+b) will be prime if a=b=2 or a=b=5 (in all case when a=b=prime).
_________________

8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Answer: B.

Theory: Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.

Hi Bunuel Quick question on Statement 1 Here Let us assume SET S with 10 elements and let 5 th and 6th element be 1/a and 1/b where a and b are both prime. Now the median of the set would be a+b/2ab and it is given that reciprocal of the median is also prime.which means that 2(ab/a+b) will also be prime. So we have to achieve this ab=a+b and that is only possible when a and b are equal to 2.And this proves that Statement 1 is sufficient to answer the question. Can you please point where did I went wrong.

2ab/(a+b) will be prime if a=b=2 or a=b=5 (in all case when a=b=prime).

Oh Yes, I missed 5. I have one more question.I tend to make these kinda mistakes in DS questions and I feel so miserable that I think it through and in the end choose the wrong answer. Any suggestions for me to improve upon these kinda stupid mistakes.

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Hi Bunuel,

Thanks for your questions, you're the Benchmark.

In this particular question I was wondering if we are assuming the S is a set of consecutive integers. Let me explain, If S is [1,-2,3] C will be insufficient. So what I'm missing?

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

Hi Bunuel,

Thanks for your questions, you're the Benchmark.

In this particular question I was wondering if we are assuming the S is a set of consecutive integers. Let me explain, If S is [1,-2,3] C will be insufficient. So what I'm missing?

Thanks

Your set {-3, -2, 1} does not satisfy any of the two statements:

(1) The product of any three integers in the set is negative --> the product of the three terms of your set is positive: (-3)(-2)1=6.

(2) The product of the smallest and largest integers in the set is a prime number --> the product of the smallest and largest integers in your set is NOT a prime: (-3)1=-3.

The next set of medium/hard DS number properties questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers.

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number (2) \(2\sqrt{x^2}\) is a prime number.

The next set of medium/hard DS number properties questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers.

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number (2) \(2\sqrt{x^2}\) is a prime number.

Could you please explain question 5 on the part "Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. "?

Could you please explain question 5 on the part "Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. "?

Thanks

I think you are talking about questions 7.

For (1): \(x=2*3*(2q+1)=2*odd*odd\) --> two odd multiples there obviously do not have 2 as their factors, thus x has 2 in power of 1 only (we have only one 2 there).

Hello, for question 9, could you pls explain how the maximum value of [a]+[b] =-2? I dont get it...I assumed a=-1 so [a] = -1, and b=-2, so [b]=-2...that means ab=2...but [a]+[b]= -1-2=-3...how do I get a -2? (as a maximum value if both are negative). thank you

gmatclubot

Re: New Set: Number Properties!!!
[#permalink]
10 May 2014, 10:47

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